[noiseordinance]

Hello there,

I'm answering a few questions for a lab and am stuck. Here's the question. "The pain reliever phenacetin is soluble in cold water to the extent of 1.0 g/1310 mL and soluble in diethyl ether to the extent of 1.0 g/90 mL.

a) Determine the approximate distribution coefficient for phenacetin in these two solvents.
b) If 50 mg of phenacetin were dissolved in 100 mL of water, how much ether would be required to extract 90% of the phenacetin in a single extraction?
c) What percent of the phenacetin would be extracted from the aqueous solution in b) by two 25-mL portions of ether?

I believe I have a) solved. K = [phenacetin in di. ether]/[phenacetin in water] = (1.1 / 100mL)/(7.6x10^-2 / 100mL) = 15.

I'm stuck on b), however. This has gotta be easy and I'm just not seeing it. Can anyone offer any direction? I imagine I can start out by figuring out how much diethyl ether is required to extract 100% first, but I don't even know how to get at that...

[Babcock_Hall]

I just have time for a quick reply right now, but your idea about calculating what it would take for 100% extraction does not look promising.  This is basically an equilibrium problem, with 90% in one phase and 10% in another.

[noiseordinance]

I just have time for a quick reply right now, but your idea about calculating what it would take for 100% extraction does not look promising.  This is basically an equilibrium problem, with 90% in one phase and 10% in another.

Thanks for the reply. This is one idea I've had... can someone verify?

K = [solv 2]/[solv 1] = (0.045g / x mL)/(0.005g / 100mL), x = 60mL

Any thoughts?

[Dan]

60 mL looks good to me

[noiseordinance]

Sweetness. Thanks much!!